Suppose that a function is twice differentiable and both are continuous, also suppose that , show that such that .

Proof so far.

By the Taylor's Theorem, we know that where is a number between 0 and x.

So we have

Let , then .

Is this right? Thanks.

Dec 3rd 2008, 09:16 AM

Opalg

There's something wrong with this question, if the inequality is supposed to hold for all . For example, the function satisfies the hypotheses of the question but not the conclusion.

The best you can realistically ask for is that there exists some neighbourhood of the origin, say , such that whenever .

With that amendment to the question, your proof is partially correct. But there's a defect in it. The problem is that your choice of depends on x. You need to adjust the proof so as to get an estimate of the size of that does not depend on x. This is possible because you are told that f'' is continuous. That means that it is bounded in any closed bounded interval. So given the interval there exists M>0 such that for all x in that interval. You can then use M in place of in your proof, define , and the proof will work nicely.

Dec 3rd 2008, 10:35 AM

tttcomrader

Thanks for the help.

The second part of this question is to find an exmaple of a differentiable function such that , but yet no inequality of the form holds in any neighborhood of the origin.

So I guess would be the example then?

Dec 3rd 2008, 11:16 AM

Opalg

Quote:

Originally Posted by tttcomrader

Thanks for the help.

The second part of this question is to find an exmaple of a differentiable function such that , but yet no inequality of the form holds in any neighborhood of the origin.

So I guess would be the example then?

No! – because does satisfy an inequality of that form in a neighbourhood of the origin. It just doesn't satisfy such an inequality globally, which is what you were previously asking for.

To get the right sort of example here, you need to ask how the proof of the first part of the question could go wrong. What you should notice is that it depended crucially on f''(x) being bounded in a neighbourhood of the origin. So you need to look for a candidate for f''(x) that is unbounded near the origin. But it must have a continuous integral f'(x). The sort of function that might do the job is . Try integrating that twice and see if it gives a function f(x) with the required properties.

Dec 4th 2008, 07:14 AM

tttcomrader

Claim: satisfies the hypothesises but the inequality do not hold in any neighborhood of the origin.

Proof so far.

Now, , so

I need to show that in any neighborhood of the origin,

Given and for some , we have

But how would I get a positive number on the right hand side? Thanks.

Dec 4th 2008, 09:18 AM

Opalg

Quote:

Originally Posted by tttcomrader

Claim: satisfies the hypothesises but the inequality do not hold in any neighborhood of the origin.

Proof so far.

Now, , so

I need to show that in any neighborhood of the origin,

Given and for some , we have

But how would I get a positive number on the right hand side? Thanks.

Do this by contradiction. Suppose that there is a neighbourhood U of 0, and a constant c>0, such that for all x in U. Then in particular that would hold for all positive x in U, so that whenever . Square both sides, divide by x^3, and you see that . But since U is a neighbourhood of 0 it must certainly contain points x satisfying . That contradiction shows that no such neighbourhood U and constant c can exist.